ESERCITAZIONE 3 HUMAN GENETICS BLOOD GROUPS RECOGNITION OF

ESERCITAZIONE 3 HUMAN GENETICS BLOOD GROUPS RECOGNITION OF PATHERNITY FAMILY PEDIGREE

INCOMPLETE DOMINANCE AND CO-DOMINANCE INCOMPLETE DOMINANCE: Is a form of intermediate inheritance in which one allele for a specific trait is not completely expressed over its paired allele. This results in a third phenotype in which the expressed physical trait is a combination of the phenotype of both alleles. (ex: color of Bella di notte petals ) CO-DOMINANCE: In co-dominance an additional phenotype is produced, however both alleles are expressed completely. The eterozygous shows the traits of both parents. ( ex: blood group)

1. The human blood groups system ABO is regulated by 3 alleles according to the following scheme: BLOOD GROUP(Phenotype) A Possible Genotypes B IB IB IB i IA IB ii AB 0 IA IA IA i Which is the dominance relation in this series of multiple alleles? IA is dominant over i IB is dominant over i IA and IB are codominant

2. Determine parental genotypes in the following crosses: Phenotypes of crossed individuals Phenotypes in the progeny a) A X A ¾ A; ¼ 0 b) A X AB A; AB c) B X 0 B; 0 d) B X A B; A; 0; AB Genotypes of crossed individuals IA i X IA i a) The parents have one dominant allele IA therefore the parents could be IA IA o IA i. In the progeny we have the phenotype 0 (recessive trait) then the parents must be IA i Frequencies of expected genotypes in the progeny : IA IA(1/4), IA i (2/4), ii (1/4) Frequencies of phenotypes: A (3/4) e O (1/4)

2. Determine parental genotypes in the following crosses : Phenotypes of crossed individuals Phenotypes in the progeny a) A X A A; 0 b) A X AB ½ A; ½ AB c) B X 0 B; 0 d) B X A B; A; 0; AB Genotypes of crossed individuals IA IA X IA IB b) The second parent must be IA IB and his gametes will be (½) IA and (½) IB The first parent could be IA IA o IA i but since we don’t have B phenotype in the progeny then he must be homozygous IA IA

2. Determine parental genotypes in the following crosses : Phenotypes of crossed individuals Phenotypes in the progeny a) A X A A; 0 b) A X AB A; AB c) B X 0 ½ B; ½ 0 d) B X A B; A; 0; AB Genotypes of crossed individuals IB i X ii c) The second parent is homozygous recessive ii, while the first one could be IB IB / IB i Since in the progeny individuals homozygous recessive appear (0 group), the first parent must be heterozygous IB i.

2. Determine parental genotypes in the following crosses : Phenotypes of crossed individuals Phenotypes in the progeny a) A X A A; 0 b) A X AB A; AB c) B X 0 B; 0 d) B X A ¼ B; ¼ A; ¼ 0; ¼ AB Genotypes of crossed individuals IB i X IA i d) Parent Genotypes: IB IB / IB i x IA IA / IA i i In the progeny we have individuals homozygous recessive ii (group 0) then the first parent must be heterozygous IB i and also the second one must be heterozygous IA i

3. Rh system is determined by the gene D: the presence of antigen (Rh+ individual) is due to the dominant allele D; the absence of antigen (Rh – individual) is homozygote for the recessive allele d. Determine the genotypes of individuals crossed: Phenotypes of crossed individuals Phenotypes in the progeny AB Rh+ X 0 Rh+ 3/8 A Rh+; 3/8 B Rh+; 1/8 A Rh-; 1/8 B Rh- X A Rh+ 1/4 AB Rh+; 1/4 A Rh+; 1/4 B Rh+; 1/4 0 Rh+ Genotypes of crossed individuals IA IB Dd x ii Dd a) The first parent must be IA IB , while the second one is ii. b) Factor Rh: both parents could be DD/Dd. Since we have individuals with Rh- in the progeny, the parents must be heterozygous Dd

Now use the brunch diagram to predict the expected phenotypes classes and their frequencies in progeny. IA IB Dd x i i Dd Gene I IA IB x i i A (1/2) B (1/2) Gene D Dd x Dd Expected phenotypes Rh+ (3/4) A Rh+ (1/2 x 3/4 = 3/8) Rh- (1/4) Rh+ (3/4) A Rh- (1/2 x 1/4 = 1/8) B Rh+ (1/2 x 3/4 = 3/8) Rh- (1/4) B Rh- (1/2 x 1/4 = 1/8)

3. Rh system is determined by the gene D: the presence of antigen (Rh+ individual ) is due to the dominant allele D; the absence of antigen (Rh – individual) is homozygote for the recessive allele d. Phenotype of crossed individuals Offsprings phenotypes Genotypes of parents AB Rh+ X 0 Rh+ 3/8 A Rh+; 3/8 B Rh+; 1/8 A Rh-; 1/8 B Rh- X A Rh+ 1/4 AB Rh+; 1/4 A Rh+; 1/4 B Rh+; 1/4 0 Rh+ IB i dd x IA i DD a) The genotype of first parent could be IB IB o IB i (blood group) and dd for Rh factor. The second parent could be IA IA o IAi and DD o Dd. b) Since in the progeny the recessive trait 0 appears but Rh- does not emerge, both parents must have one recessive allele about blood group then the genotype will be IB i x IA i. About Rh, the second parent must be DD 4 classes of phenotypes are expected with the same frequency (1: 1: 1: 1)

Now use the brunch diagram to predict the expected phenotypes classes and their frequencies in the offspring IB i d d x I A i D D Gene I Gene D IB i x IA i dd x DD Expected phenotypes A (1/4) B (1/4) AB (1/4) O (1/4) A Rh+ (1/4) B Rh+ (1/4) AB Rh+ (1/4) O Rh+ (1/4) Rh+ (1)

4 - The human blood groups system ABO is regulated by 3 alleles IA, IB, i. Rh system is determined by the gene D. The group MN is control by L gene with LM and LN alleles (codominant alleles). These three genes are independent. Which gametes and with which frequencies will be produce by these genotypes? Genotypes Gametes a) IBIB Dd LNLN IB D LN ½ IB d LN ½ B Rh+ N b) IBi Dd LMLN IB IB 1/8 , i D LM D LN d LM d LN B Rh+ MN c) IAi Dd LMLM IA D LM ¼ , i D L M ¼ IA d LM ¼ , i d L M ¼ A Rh+ M d) IAIB dd LMLN IA d LM ¼, IA d LN ¼, AB Rh- MN e) ii dd LMLN i d LM ½ , i d L N ½ D LM D LN d LM d LN Phenotype I B d LM ¼ I B d LN ¼ 1/8 1/8 O Rh- MN

4. Mother and son have the indicated blood groups. Could the man be the father of the child? Mother’s genotype: IA IA / IA i dd child’s genotype: IB IB / IB i DD/Dd man’s genotype: IB IB / IB i dd The man is not the father of the child. He hasn’t the allele D.

4. Mother and son have the indicated blood groups. Could the man be the father of the child? 0 Rh- 0 Rh+ A Rh+ Mother’s genotype: IA IA / IA i DD/Dd child’s genotype: i i DD/Dd man’s genotype: i i dd The man could be the father of the child if the mother’s genotype is IA i DD/Dd

4. Mother and son have the indicated blood groups. Could the man be the father of the child? AB Rh+ A Rh- 0 Rh- Mother’s genotype: i i dd child’s genotype: IA IA / IA i dd man’s genotype: IA IB DD/Dd The man could be the father of the child if his genotype is Dd

5. Figure out the probability of these events. Mother’s genotype: IA IB dd Father’s genotype: i i DD/Dd child’s genotype: IB i dd The child is certainly heterozygous for blood group because he inherited ì allele from the father. Since the child is dd, the father must be heterozygous Dd The probability that the cross IA IB dd x i i Dd produce a child IB i dd is: Probability of phenotype B = ½ x 1 = ½ Probability of phenotype Rh- = 1 x ½ = ½ Probability of phenotype B Rh- = ½ x ½ = 1/4

5. Figure out the probability of these events. Mother’s genotype: Father’s genotype: child’s genotype: i i DD/Dd IA IA / IA i DD/Dd i i dd Since the child is homozygous recessive for both genes, the father must be heterozygous IA i Dd. The mother must be heterozygous for Rh factor: ii Dd The probability that the cross ii Dd x IAi Dd produce a child ii dd is: Probability of phenotype 0 = ½ x 1 = ½ Probability of phenotype Rh- = ½ x ½ =¼ Probability of phenotype 0 Rh- = ½ x ¼ = 1/8

Exercise 1 B Rh+ N O Rh- MN O Rh- N -probability of this event -genotypes of parents ½ x 1/2 =1/8 Ibi Dd LNLN ii dd LMLN Child Phenotype Parent Phenotypes

Exercise 2 AB Rh+ MN B Rh- MN A Rh- M Child Phenotype -probability of this event -genotypes of parents and child 1/4 x ½ x 1/4 =1/32 Ib Ia Dd LMLN Ib i dd LMLN Ia i dd LMLM Parent Phenotypes

1 - In family trees the most used symbols are : Male mutant Female normal I Parents II children 1 2 3 A roman number is assigned to every generation and the individuals of the same generation are progressively numbered from left to right with Arabic numbers.

INHERITANCE DOMINANT AUTOSOMAL Male and female are usually affected with equal frequency. DOMINANT All the generations have individuals that show the trait Affected parents can have unaffected children RECESSIVE The affected individuals are rare and they are not present in all the generations. Unaffected parents can have affected children RECESSIVE SEX-LINKED DOMINANT Disease never transfer from father to son Daughters of an affected father will be affected RECESSIVE Males are more affected Disease never transfers from father to son Disease tends to transfer from mother to son

In the following family trees, black symbols represent a mutant phenotype. Determine if the mutant phenotype is due to a dominant or a recessive allele, taking into consideration that the mutant allele is rare A- aa Aa Aa aa aa Aa Mutant phenotype is present in all generations. Mutants derive from almost one mutant parent The mutant allele is DOMINANT The mutant individuals are heterozygous or homozygous dominant (AA o Aa) while the normal individuals are homozygous recessive (aa).

In the following family trees, black symbols represent a mutant phenotype. Determine if the mutant phenotype is due to a dominant or a recessive allele, taking into consideration that the mutant allele is rare Aa aa Here we have mutant sons from normal individuals. The mutated allele is RECESSIVE. The mutant individuals are homozygous recessive, the parents are heterozygous.

Considering people that become part of the family through marriage don’t have the mutant allele, calculate the probability that an affected (sick) child is born from cross III, 3 and III, 4 Aa Aa AA AA aa aa Aa (1) A a A AA 1/4 Aa 1/4 aa 1/4 A AA 1/4 a Aa 1/4 Aa (2/3) ? Aa (½) Female III, 3 is heterozygous Aa (probability=1) because she is normal and the mother is mutant (homozygous recessive) Man III, 4 could be heterozygous if his father is heterozygous (II, 5). The probability that II, 5 is heterozygous is 2/3 (he is normal son of heterozygous parents). The probability that III, 4 inherits the recessive allele is ½ (the cross is Aa X AA (external)). The probability that an affected (sick) child is born from cross III, 3 and III, 4 is ¼. The probability that all these events happen is the product of all the probabilities: 1 x 2/3 x ½ x ¼ = 1/12

8. in the following family trees, black symbols represent the mutant phenotype of recessive homozygote. Considering that mutant alleles are rare, calculate the probability of an affected child in the indicated crossed. a) I II AA (1) aa (1) Sick individuals of II generation are homozygous recessive with probability 1 III Aa IV Aa (1) ? Aa (1/2 ) (1/4 ) The individuals of IV generation could produce a sick child if they are heterozygous. The probability that IV, 1 is heterozygous is 1 because the cross is AA x aa Probability that IV, 2 is heterozygous? ? !!!. III, 4 is heterozygous (p=1) because the cross is aa x AA and he donates his recessive allele. The son IV, 2 possess the recessive allele with the probability of ½ Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4 The probability that all these events happen is the product of all the probabilities : 1 x ½ x ¼ = 1/8

8. in the following family trees, black symbols represent the mutant phenotype of recessive homozygote. Considering that mutant alleles are rare, calculate the probability of an affected child in the indicated crossed. aa Aa (1) III, 3 e III, 4 must be both heterozygous in order to produce a sick son. Aa (1/2) (1/4) III, 3 is certainly heterozygous because his mother is homozygous recessive (p=1) III, 4 will be heterozygous if he inherited from his father the recessive allele. The probability that the father II, 3 is heterozygous is 1 (the parents are aa x Aa) III, 4 is heterozygous with probability ½. Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4 The probability that all these events happen is the product of all the probabilities : 1 x ½ x ¼ = 1/8

8. in the following family trees, black symbols represent the mutant phenotype of recessive homozygote. Considering that mutant alleles are rare, calculate the probability of an affected child in the indicated crossed. Aa Aa AA AA aa (2/3) Aa Aa (1) III, 1 and III, 2 must be both heterozygous to have a sick son. Aa (1/2) III, 1 is certainly heterozygous (p=1). III, 2 could be heterozygous if the father is heterozygous. The probability that II, 3 is heterozygous is 2/3 (health son of the heterozygous parents) III, 2 is heterozygous with probability of ½. Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4 The probability that all these events happen is the product of all the probabilities: 1 x 2/3 x 1/2 x ¼ = 1/12

8. in the following family trees, black symbols represent the mutant phenotype of recessive homozygote. Considering that mutant alleles are rare, calculate the probability of an affected child in the indicated crossed. aa (1) Aa (2/3) Aa (1) II, 3 will be heterozigous with P=2/3, (health son of heterozygous parents) II, 4 is heterozygous p=1 since her mother is homozygous recessive. Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4 The probability that all these events happen is the product of all the probabilities: 2/3 x 1 x ¼ = 2/12=1/6

9 - If individuals of cross II 3 -II 4 have already had an affected child, which is the probability to have a second affected child? If these parents already have a sick child, we are sure that they must be heterozygous Aa (1). Then the probability to have a second affected child is ¼. aa (1) Aa (1) Probability that starting from the cross Aa X Aa we have a sick child (aa) = 1/4

Aa 1 aa 1 Aa 1 aa 1 AA 1 Aa 2/3 aa 1/4 x 2/3=1/6

. In the following family trees, black symbols represent the mutant phenotype of recessive homozygote. Considering that mutant alleles are rare, calculate the probability of an affected child in the indicated crossed. Considering people that become part of the family through marriage don’t have the mutant allele. A- aa I Aa AA II ? aa P=0 Aa Aa I aa AA II ? aa P=0

. In the following family trees, black symbols represent the mutant phenotype of recessive homozygote. Considering that mutant alleles are rare, calculate the probability of an affected child in the indicated crossed. Considering people that become part of the family through marriage don’t have the mutant allele. aa A- I Aa aa II ? aa P=1/2 Aa I Aa Aa 2/3 II P=0 AA

. In the following family trees, black symbols represent the mutant phenotype of recessive homozygote. Considering that mutant alleles are rare, calculate the probability of an affected child in the indicated crossed. Considering people that become part of the family through marriage don’t have the mutant allele. Aa Aa I Aa 2/3 aa II aa 1/2 P=2/3 x 1/2=1/3 I A- AA Aa II III aa Aa Aa AA aa Aa Aa 1/2 aa 1/4 P=1/4 x 1/2=1/8

Aa I II III aa Aa Aa Aa 2/3 Aa AA aa AA Aa Aa 1/2 aa 1/4 P=1/2 x 1/4 x 2/3=1/12

Scheda 4. SEX LINKAGE

Human Female Karyotype Human Male Karyotype X-LINKED inheritance: If the trait is recessive and associated with X chromosome, the phenotype will be more frequent in males (XY) than in females (XX) since they are HEMIZYGOTE.

1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PROGENY PHENOTYPES PARENTS PHENOTYPES FEMALE pale MALE Xb Xb dark FEMALES XB Y dark 45 pale 0 MALES XB Xb dark 0 pale 48 Xb Y a) PARENTS HAVE DIFFERENT PHENOTYPES. b) IN THE PROGENY THE PHENOTYPEs DISTRIBUTION is NOT EQUAL BETWEEN MALES AND FEMALES. c) THE GENE IS X-LINKED d) In order to establish which is the dominant allele, i observe the females: they are dark then they are XBe)The males are hemizygote then the phenotype corresponds to the unique allele that they carried. The dark parent will be B then XB Y, while the pale sons will be b (recessive allele inherited from the mother) then Xb Y f) The mother is homozygous recessive Xb Xb while the dark daughters will be heterozygous XB (father’s allele) Xb (mother’s allele)

1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PROGENY PHENOTYPES PARENTS PHENOTYPES Female rough Male Rr rough Females Rr Rough 87 Smooth 33 Males RR/Rr Rough 92 RR/Rr Smooth 27 r r rr a) The parents have the same phenotype b) In the progeny males and females have the same distribution. (1: 3) c) The gene is autosomic. d) Rough is dominant (R), the recessive (r) determine the smooth phenotype: The parents are both heterozygous. (Rr). e) From Rr X Rr, we expect ¾ rough (1/4 RR + 2/4 Rr) and ¼ smooth (1/4 rr) If the gene was x-linked? All the females will be rough because they inherit the father’s allele R.

1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PROGENY PHENOTYPES PARENTS PHENOTYPES Female red Male XR Xr red Females XR Y Red 102 White 0 Males XR - Red 49 XR Y White 52 Xr Y a) The parents have both the same phenotype b) In the progeny we have non equal distribution between males and females. c) The gene is x-linked d) Dominant allele? = females phenotypes (all red individuals then Red is dominant (X R) e) The males are red and white. The sons will be: red XR Y and white Xr Y f) The mother’s genotype must be XR Xr: the sons have two different genotypes (XR Y, Xr Y) g) The daughters have two different genotypes (XR XR e XR Xr ) but all of them are red.

1. Determine for each cross if the character is sex-linked or not and assign the genotype to the crossed individuals. PROGENY PHENOTYPES PARENT PHENOTYPES Female black Male NN white Females nn black 98 Males Nn withe 0 black 103 N n white 0 a) The parents have different phenotypes. b) All the individuals of the progeny are black, black is the dominant character (N). c) Since the trait does not segregate, we cannot define if the gene is autosomal or xlinked d) The mother is homozygous dominant: N N or XN XN e) The father is homozygous recessive: n n or Xn Y PROGENY PHENOTYPES PARENT PHENOTYPES Female black Male XN XN white Females Xn Y black 98 white 0 Males XN Xn black 103 XN Y white 0

2. In humans the presence of a fissure in the iris (coloboma iridis) is controlled by a recessive sex-linked gene. An affected daughter is born from a normal couple. The husband asks for divorce, accusing his wife of infidelity. Is he right? We can draw the family tree. XA Y XA Xa Xa Xa The daughter must be homozygous recessive. Then the parents contribute with gametes Xa The mother must be heterozygote because she is health. Yes he is right he can not be the father of the girl.

A recessive sex-linked gene is responsible for the most common kind of haemophilia. Considering the family tree shown below, answer the following questions: XE Y XE Xe I II Xe Y The mother gave the mutant allele and she is heterozygous XE Xe, while the father will be XE Y The son II, 3 is Xe Y. The daughter II, 2 inherited the allele Xe with probability 1/2 (½ XE XE, ½ XE Xe). a) If II. 2 marries a normal man (XE Y), which is the probability to have children affected by haemophilia? The probability to produce sick child will ¼ x ½ = 1/8 (haemophilic males). b) If II. 2 already has an affected child, which is the probability to have a second affected child? If the II. 2 has already a sick child, the mother is certainly heterozygous then the probability is ¼ x 1= ¼.

Determine if the character shown in the following family tree can be due to: a) the dominant allele of an autosomic gene aa Aa Aa Aa aa aa aa Try to complete the scheme with the genotypes. If the allele is dominant, all the healthy individuals will be homozygous recessive aa. The sick individuals will be homozygous dominant AA or heterozygous Aa. The individual I, 1 is heterozygous because in the second generation not all the individuals are sick. Hypothesis could be right

4. b)the recessive allele of an autosomal gene Aa aa aa aa Aa Aa aa Aa A- A- If the allele is recessive, all the sick individuals are homozygous recessive aa. If the character is autosomal recessive we have to suppose that two individuals that come in to the family with marriage are heterozygous. If the character is rare this is not probable Hypothesis could not be right

c) the dominant allele of a sex-linked gene; XA Y Xa Y XA Y Xa Xa XA Xa Xa Y XA Xa XA Y Xa Xa Xa Y Xa Xa If the allele is x-linked and dominant, all the healthy females are homozygous recessive Xa Xa and the healthy males are Xa Y. The sick men will be XA Y. The sick females will be heterozygous XA Xa. Hypothesis could be right

d) the recessive allele of a sex- linked gene Xa Y XA Y Xa Y XA Xa Xa Xa XA Y Xa Xa Xa Y XA Xa XA Y XA XA - If the allele is x-linked and recessive, all the sick males are hemizygote X a Y All the health males will be hemizygote XA Y. The sick females of the second generations must be homozygous Xa Xa then the original mother must be heterozygous XA Xa Hypothesis could be right, but if the character is rare this configuration is not probable.